Consider the following bilinear saddle point problem: \begin{align*} \min_x \max_y &~ f(x,y)\\ \text{where} &~ f(x,y) = x^\top A y + b^\top x + c^\top y, ~\text{and}~ x,y,b,c \in \mathbb{R}^d, A \in \mathbb{R}^{d \times d}. \end{align*}
Suppose this problem has unique Nash Equilibrium, and denote it by $(x^\ast,y^\ast)$. Now we regularize this objective by \begin{align*} \tilde{f}(x,y) = f(x,y) + \alpha \cdot g(x), \end{align*} where $g: \mathbb{R}^d \mapsto \mathbb{R}$ is a smooth and strongly-convex function and $\alpha \in \mathbb{R}$. Let $(\tilde{x}^\ast,\tilde{y}^\ast)$ denote the Nash Equilibrium of the regularized mini-max problem, i.e. \begin{align*} \tilde{x}^\ast = &~ \arg \min_x \max_y \tilde{f}(x,y)\\ \tilde{y}^\ast = &~ \arg \max_y \tilde{f}( \tilde{x}^\ast,y). \end{align*}
The Questionis: Does $\|x^\ast - \tilde{x}^\ast\|_2 = O(\alpha)$ (say $\alpha \rightarrow 0$) hold? If not, is there any counterexample?
$\|x^\ast - \tilde{x}^\ast\|_2 = O(\alpha)$ looks correct to me, but I tried and didn't find any proof. Great thanks if anyone can drop some clues!